We are given two logarithm problems to solve for $x$. Problem 5: Find $x$ if $\log_{10}(2x+1) - \log_{10}(3x-1) = 1$. Problem 6: Find $x$ if $\log_{10}5 + \log_{10}(x+2) - \log_{10}(x-1) = 2$.

1. Problem Description

We are given two logarithm problems to solve for $x$.

Problem 5: Find $x$ if $\log_{10}(2x+1) - \log_{10}(3x-1) = 1$.

Problem 6: Find $x$ if $\log_{10}5 + \log_{10}(x+2) - \log_{10}(x-1) = 2$.

2. Solution Steps

Problem 5:

$\log_{10}(2x+1) - \log_{10}(3x-1) = 1$

Using the logarithm property $\log_a(b) - \log_a(c) = \log_a(\frac{b}{c})$, we have:

$\log_{10}(\frac{2x+1}{3x-1}) = 1$

Convert the logarithmic equation to an exponential equation using $log_a(b)=c \implies a^c = b$:

$\frac{2x+1}{3x-1} = 10^1$

$\frac{2x+1}{3x-1} = 10$

Multiply both sides by $(3x-1)$:

$2x+1 = 10(3x-1)$

$2x+1 = 30x - 10$

Subtract $2x$ from both sides:

$1 = 28x - 10$

Add $10$ to both sides:

$11 = 28x$

Divide both sides by $28$:

$x = \frac{11}{28}$

We need to check if this value of $x$ is valid by plugging it back into the original equation.

$2x+1 = 2(\frac{11}{28}) + 1 = \frac{11}{14} + 1 = \frac{25}{14} > 0$

$3x-1 = 3(\frac{11}{28}) - 1 = \frac{33}{28} - 1 = \frac{5}{28} > 0$

Since both $2x+1$ and $3x-1$ are positive for $x = \frac{11}{28}$, this solution is valid.

Problem 6:

$\log_{10}5 + \log_{10}(x+2) - \log_{10}(x-1) = 2$

Using the logarithm property $\log_a(b) + \log_a(c) = \log_a(bc)$ and $\log_a(b) - \log_a(c) = \log_a(\frac{b}{c})$, we have:

$\log_{10}(\frac{5(x+2)}{x-1}) = 2$

Convert the logarithmic equation to an exponential equation using $log_a(b)=c \implies a^c = b$:

$\frac{5(x+2)}{x-1} = 10^2$

$\frac{5x+10}{x-1} = 100$

Multiply both sides by $(x-1)$:

$5x+10 = 100(x-1)$

$5x+10 = 100x - 100$

Subtract $5x$ from both sides:

$10 = 95x - 100$

Add $100$ to both sides:

$110 = 95x$

Divide both sides by $95$:

$x = \frac{110}{95} = \frac{22}{19}$

We need to check if this value of $x$ is valid by plugging it back into the original equation.

$x+2 = \frac{22}{19} + 2 = \frac{22}{19} + \frac{38}{19} = \frac{60}{19} > 0$

$x-1 = \frac{22}{19} - 1 = \frac{22}{19} - \frac{19}{19} = \frac{3}{19} > 0$

Since both $x+2$ and $x-1$ are positive for $x = \frac{22}{19}$, this solution is valid.

3. Final Answer

For problem 5: $x = \frac{11}{28}$

For problem 6: $x = \frac{22}{19}$